If $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is :

$\frac{-2}{1+x^2}$ $\frac{2}{1+x^2}$ $\frac{1}{2+x^2}$ $\frac{2}{2-x^2}$

$\frac{-2}{1+x^2}$

Put x = $\tan \theta$ ∴  $y=\sin ^{-1}(\cos 2 \theta)=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2 \theta\right)\right)$ $=\frac{\pi}{2}-2 \theta=\frac{\pi}{2}-2 \tan ^{-1} x$ $\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^2}$ Hence (1) is correct answer.